Tell me more ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.
up vote 2 down vote favorite
1

How can I interpret the parameter temperature $T$, if I'm not given the description of the system in terms of the equation of state, $E(S,V\ )$ or $S(E,V\ )$ and so on.

In many systems it makes sense to think of it as an "energy" itself, e.g. when the entropy is such, that $T$ basically represents the mean kinetic energy of particles. The general definition goes like $$\frac 1 T=\small{\left(\frac{\partial S(E,V\ )}{\partial E}\right)_V}.$$

However, very often, especially when I'm reading about phase transitions, examples for systems with critical exponents and so on, they usually talk about the parameter $T$ and an associated critical $T_c$ without taking any specific system into accound. There is usually some abstract free energy $F$, which pops out abstract quantities. Or in the Boltzman distribution and derived quantities, often something gets activated when the value of $kT$ catches up with some system specific energy value $E_0$. And it gets more complicated when it takes QFT like form.

What is the temperature in a general setting.

How do I read this kind of things and what should I have in mind when reading these kinds of texts?

share|improve this question
1  
You wrote the answer. This is the definition of $T$. In the Ising model. for example, there is no sense in talking about "average kinetic energy of the spins" or anything of that sort. I'd say that a good way to think about the temperature in this case is "how far above the ground state can I go", which is roughly equivalent to "how much energy can I pay in order to buy some entropy"? – yohBS Jan 16 '12 at 21:03
Could you elaborate on the sentences in parentheses? Why "can I go"? Can I go with what? I have a probability for all energy levels to be to taken. And why "pay"? Why "buy" entropy? Do I want entropy? – Nick Kidman Jan 17 '12 at 8:36
1  
Systems flow towards probable states, that is - states that have a large number of micro-states. The (log of the) number of micro states is the entropy. So systems "want" to be in (="go to") states with high entropy, but that usually means states with high energy. The trade-off between entropy and energy is exactly the temperature. In evaporating liquid, for example, the vapor state has a higher entropy, but also a higher energy. Therefore, at low $T$ the system is liquid, but when $T$ is high enough the system prefers to be at a higher energetic state (="pay") because its entropy is higher. – yohBS Jan 17 '12 at 12:00
@yohBS: Your explanation is formulated in terms of the system, which is not in equilibrium. Would you say one should view and describe the meaning of the temperature $T$ functionally, i.e. in that its the value, which equals out for coupled subsystems, if the whole system reaches equilibrium? – Nick Kidman Jan 17 '12 at 17:10
I posted a real, elaborate answer. Hope that gets the job done. – yohBS Jan 17 '12 at 21:20
show 3 more comments

3 Answers

up vote 3 down vote accepted

The most fundamental definition of temperature is derived from the zeroth law of thermodynamics.

The zeroth law declares thermal equilibrium an equivalence relationship, and thus we can tag each equivalence class with a number that we call temperature. Or in less mathematical term, temperature is a physical quantity tagged to each thermodynamic system such that any two systems with the same temperature would stay in thermal equilibrium when they contact.

The exact way of assigning temperature to a system is called a temperature scale. There were multiple scales before, most based on thermal properties of a particular substance. Then Kelvin devised a scale based solely on thermodynamic principles, which we call "absolute scale".

share|improve this answer
Interesting points! I wonder if there zeroth law can't be deduced within thermodynamics, do you (if no) know why not? Regarding your last paragraph, I think the Kelvin definition and the absolute isn't necessary if (via statistical mechanics) you can compute the absolute entropy. – Nick Kidman Jan 19 '12 at 11:18
@Nick Kidman: The zeroth law is "zeroth" because it was formulated after the first and second law when people realized it is implicit in their understanding, yet an independent and fundamental law. So no, you cannot deduce it within thermodynamics. As for "absolute scale", or thermodynamical scale, there are multiple definitions, all equivalent. Kelvin's version is the first one, though. – C.R. Jan 20 '12 at 2:13
up vote 3 down vote

Stat Mech is all about taking averages with the proper measure.

The simplest measure is the micro-canonical measure (ensemble), in which you assume that the system has a given energy, and that all the states in this energy shell are equally probable. The assumption that the energy is known basically mean that the system does not exchange energy with a bath.

A much more useful measure is the canonical measure, in which you assume that the average energy is known, but that the system may exchange energy with a heat bath. It can be shown (and it is shown here) that if you assume that in each energy shell the states are equi-probable, this probability is proportional to $e^{-E/k_bT}$.

This shows you that $T$ measures the weight that is given to states, according to their energy. When $T$ is high, the weight of a state with a given energy becomes higher, and when $T$ is low it becomes lower. It is easily seen that for $T\to\infty$ all the states become equi-probable and for $T\to 0$ only the ground state(s) is (are) counted.

You might ask then, "so why isn't the system almost always at the ground state? It is the most probable one!". Here entropy goes into play. By definition of the entropy $S(E)$, the number of states with an energy $E$ is $e^{S(E)/k_b}$, and it is a rapidly increasing function of $E$. Therefore, when you average over energies you should use the measure $$e^{-\beta E}e^{S(E)/k_b}=e^{-\beta(E-TS)}$$ because higher energies have a much larger number of states on the shell. This is what I meant when I said that $T$ is a measure of "how much energy can I pay in order to buy some entropy" - high-entropy states are more probable from state-counting considerations, but are less probable from energetic considerations. $T$ is a measure of the relative balance of these two.

Also, your suggestion of functionally defining $T$ as "the thing that is equal for two systems in thermal contact" is great, and is actually used it in this textbook which gives offers a thorough, insightful introduction to "what is temperature". I highly recommend it for beginners.

share|improve this answer
So $\int_{\text{states}} e^{-\beta E}=\int_{\text{energies}} e^{-\beta E}\Omega(E)=\int_{\text{energies}} e^{-\beta E}e^{S}=\int_{\text{energies}} e^{-\beta F}$, where $\Omega(E)$ is the phase space volume with energy $E$? I see that $E-TS$ must be minimized, but since I don't see the lower bound of this expression without knowing $E$ or $S$ I don't know how to minimize it. Also, if you're taking the canonical ensamble with a real bath, then you can't just choose $T$. In that case there is no battle between energy and entropy, since $T$ will just take the value it has to from outside. – Nick Kidman Jan 17 '12 at 21:43
The point is showing that coupling your system to an external bath results in a probability that goes like $e^{-\beta E}$, for some $\beta$. This is te definition of $T$, and is crux of the matter. Until you've shown this you don't know that "T will just take its value" from the bath, because $T$ is not even defined. – yohBS Jan 17 '12 at 21:51
forget what I wrote about f. It is confusing and wrong. In this formulation the free energy is given by $f=-k_bT\log Z=k_bT\log \sum_i e^{-\beta E_i}$, and the mean energy is calculated by averaging $\langle E\rangle=Z^{-1}\sum E_i e^{-\beta E_i}=-\partial_\beta \log(Z)$ – yohBS Jan 17 '12 at 23:00
I don't understand how "this is the definition of $T$". How do you determine $T$ from this formalism, if it's not defined as the entropy derivative with respect to $E$? And eventually do you agree that with the bath, $T$ will just take the value of the temperature of the outer bath? And what is "confusing and wrong", I don't know what you're referring to. – Nick Kidman Jan 18 '12 at 8:03
When you couple your system to a bath, you actually allow it to exchange heat with it. The total system (your system and the bath) will be in the most probable state - the one with the highest entropy. Since the total entropy is the sum of the bath's and the system's entropy, the total entropy is maximal when $$\frac{\partial S_{bath}}{\partial E}=\frac{\partial S_{system}}{\partial E}\ .$$ This quantity is defined to be $1/T$, and the probability to find your system in an energy $E$ goes like $e^{-E/kT}$, as can be seen by taking a first order Taylor series around the maximum. – yohBS Jan 18 '12 at 9:38
up vote 2 down vote

Your definition is the thermodyanmics definition of temperature. Most discussions of phase transitions are coming from a statisical mechanics point of view. In this paradigm, the definition of temperature follows from the definition of entropy $S = k_b \log \Omega(E)$ $$ \frac{1}{kT} = \frac{d \log \Omega(E)}{dE}$$ Where $\Omega(E)$ is the number of microstates with energy $E$.

Wikipedia has a nice derivation of the connection between these two approaches.

share|improve this answer
Thanks for the comment, I'm familiar with the definition. I don't know if "the multiplicative inverse of the relative phase space volume change with energy" is all too helpful though. – Nick Kidman Jan 17 '12 at 8:37

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.